Add Two Numbers

 描述

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

【题意】

给你两个链表，表示两个非负整数。数字在链表中按反序存储，例如342在链表中为2->4->3。链表每一个节点包含一个数字（0-9）。

计算这两个数字和并以链表形式返回。

思路：
思路非常简单，利用两个指针分别遍历两个链表，并且用一个变量表示是否有进位。某个链表遍历结束之后再将另一个链表连接在结果链表之后即可，若最后有进位需要添加一位。

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {        int carry = 0;        ListNode* tail = new ListNode(0);        ListNode* ptr = tail;        /**  * Definition for singly-linked list.  * struct ListNode {  *     int val;  *     ListNode *next;  *     ListNode(int x) : val(x), next(NULL) {}  * };  */ class Solution { public:     ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {         int carry = 0;         ListNode* tail = new ListNode(0);         ListNode* ptr = tail;       ```           while(l1 != NULL || l2 != NULL){             int val1 = 0;             if(l1 != NULL){                 val1 = l1->val;                 l1 = l1->next;             }             int val2 = 0;             if(l2 != NULL){                 val2 = l2->val;                 l2 = l2->next;             }             int tmp = val1 + val2 + carry;             ptr->next = new ListNode(tmp % 10);             carry = tmp / 10;             ptr = ptr->next;         }         if(carry == 1){             ptr->next = new ListNode(1);         }         return tail->next;     } };